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Show that are are only 16 integer solutions to the following equation: $$11x + 8y + 17 = xy$$ What I tried: I took a modulo 2, and I got that $y$ must be even and $x$ must be odd. But beyond that, I don't really know how to start.

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    $\begingroup$ $17=xy-11x-8y$, then $105=xy-11x-8y+88$, then $105=(x-8)(y-11)$. I will leave it for you to finish. $\endgroup$ – chubakueno Aug 17 '13 at 2:50
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Here is an algebraic method: $$17=xy-11x-8y$$ $$\iff 105=xy-11x-8y+88$$ $$\iff 105=(x-8)(y-11)$$

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    $\begingroup$ Just a note: this is not an ad-hoc trick specific to these numbers, it really does work for any $ax + by + c = xy$: we have $c = xy - ax - by$ iff $c + ab = (x-b)(y-a)$. $\endgroup$ – ShreevatsaR Aug 17 '13 at 3:14
  • $\begingroup$ @ShreevatsaR Exactly, I learnt about this method from calculating the number of integer solutions to $x^{-1}+y^{-1}=2010^{-1}$. It works pretty well in a variety of situations! $\endgroup$ – chubakueno Aug 17 '13 at 3:19
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Hint:

  1. Solve for $x$ to see, that there is an integer solution for a given $y\in\mathbb Z$, iff: $$y-11\mid 8y+17$$
  2. Conclude $y-11\mid 105$.
  3. Check the remaining (finitely many) possibilites.
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11x + 17 = y(x - 8)
y = (11x + 17)/(x - 8)
y = 11(x - 8) + {105/(x - 8)}

Now, 11(x - 8) can have infinite x. But, 105/(x - 8) should be an integer too for y to be an integer. So, we see that 105 is divisible by these numbers = 1 x 105, 3 x 35, 5 x 21, 7 x 15. So, 105 is divisible by 1,3,5,7,15,21,35,105. Also, negative of these numbers. So we have 8 + 8 = 16 numbers.

So, possible x will be solved for like this:

x - 8 = 1
x - 8 = 3
...
x - 8 = 105
x - 8 = -1
x - 8 = -3
...
x - 8 = -105

These are the values of x for which y will be an integer! (16)

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  • $\begingroup$ Factors of 105 from here: mathsisfun.com/numbers/factors-all-tool.html $\endgroup$ – Aayush Aug 17 '13 at 3:09
  • $\begingroup$ There is a formula for the number of divisors. You factor your number and then $$\text{number of divisors}=(e_1+1)(e_2+2)...(e_i+1)$$ where e_k is the exponent of the kth prime. So you don't even need to know who they are! $105=2*3*5$ implies $$\text{number of divisors}=(2)(2)(2)=8$$. However, the data you presented deserves an upvote :) $\endgroup$ – chubakueno Aug 17 '13 at 3:16
  • $\begingroup$ Yes, I know that formula but I thought that the actual data might help. Thank you! $\endgroup$ – Aayush Aug 17 '13 at 3:19

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